How do you solve #sqrt{2x + 19} = x + 8#?

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Answer 1 · 2016-09-17T05:02:43

#x=-5#

#sqrt(2x+19)=x+8#

Square both sides
#(sqrt(2x+19))^2=(x+8)^2#

Remember, to square a binomial, multiply it by itself.
Don't distribute the exponent!
#2x+19=(x+8)(x+8)#
#2x+19=x^2+8x+8x+64#
#2x+19=x^2+16x+64#

Gather all terms on one side by subtracting 2x and 19 from both sides.

#0=x^2+14x+45#

Factor
#0=(x+9)(x+5)#

Set each factor equal to zero.
#x+9=0# and #x+5=0#
#x=-9# and #x=-5#

Check for extraneous solutions by substituting each answer into the original equation.

#sqrt(2*-9+19)=-9+8#
#sqrt(1)=-1#
#1!=-1#, so #x=-9# is an extraneous (incorrect) solution

#sqrt(2*-5+19)=-5+8#
#sqrt(9)=3#
#3=3#, so #x=-5# is a solution